The equation of a circle $C$ is $x^2+y^2+6x+14y+49 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2+6x) + (y^2+14y) = -49$ $(x^2+6x+9) + (y^2+14y+49) = -49 + 9 + 49$ $(x+3)^{2} + (y+7)^{2} = 9 = 3^2$ Thus, $(h, k) = (-3, -7)$ and $r = 3$.